Quick question about Jensen inequality and Auxiliary variables

Hello all,

I have a quick question, that seems to have been discussed but I (and my colleagues) still feel puzzled a bit.

Lets assume we define two auxiliary variables Z1 and Z2 as:


using these two auxiliary variables, I am curious if dynare sees Z2(+1)as:

1-) Z2(+1)= E_t { (C_(t+1)^-(rho))*(C_(t+2)^-(rho)) } or

2-) Z2(+1)=E_t { (C_(t+1)^-(rho))*E_t { (C_(t+2)^-(rho))} }= E_t { (C_(t+1)^-(rho)) } *E_t { (C_(t+2)^-(rho))}

Thanks a lot and wish you all a nice weekend

Dynare uses 1). When building auxiliary variables when leads > 1 are involved, Dynare always deals with a entire nonlinear expression so as to respect Jensen inequality.

thanks a lot
i was worried that dynare sees Z1(+1) in Z2 as
Z1(+1)=*E_t { (C_(t+1)^-(rho)); as

See also Expected value of a power

thank you for the clarifications. very helpful. Just to be totally clear what if the auxiliary variables were (C is consumption at time t)

Z1=C(+1)^-(rho); in dynare Z1= E_t { (C_(t+1)^-(rho))}

in dynare still the same logic holds?

1-) Z2= E_t { (C_(t+1)^-(rho))(C_(t+2)^-(rho)) }
2-) Z2(+1)= E_t { (C_(t+2)^-(rho))
(C_(t+3)^-(rho)) }

or in my previous example Z1=C^-(rho); the E_t { } cancels ( as explained in Dynare timing and redefinition) hence
Z1(+2)=C(+3)^-(rho)… holds in every period without expectations so i can put Z1(+1) inside exceptions of another long complicated equation with endogenous variables without worrying about Jensen and concavity.


Are you working at order=1 or order>1? If the former, you don’t need to worry about any of this. If you are working at higher order, please use the \LaTeX capabilities to make your formulas readable. Simply use Dollar signs to start and end that code

Thanks Johannes,

I am working with risky and stochastic steady state concept and getting the right covariance matrix is essential to what i am trying to do.

  • So in dynare code if i have the following entries


these in dynare are interpreted as:

1-) Z1_{t}=\mathbb{E}_{t} [C_{t+1}^{ -\rho}]
2-) Z2_{t}=\mathbb{E}_{t} [C_{t+1}^{ -\rho}\mathbb{E}_{t} [C_{t+2}^{ -\rho}] ]=\mathbb{E}_{t} [C_{t+1}^{ -\rho}]\mathbb{E}_{t} [C_{t+2}^{ -\rho}] OR
2a-) Z2_{t}=\mathbb{E}_{t} [C_{t+1}^{ -\rho}C_{t+2}^{ -\rho} ] i believe this is wrong but not 100% sure

3-)Z1(+1) means \mathbb{E}_{t+1}[Z1_{t+1 }]=\mathbb{E}_{t+1} [C_{t+2}^{ -\rho}]

4-)Z2(+1) means \mathbb{E}_{t+1}[Z2_{t+1 }]=\mathbb{E}_{t+1}[\mathbb{E}_{t+1}[C_{t+2}^{ -\rho}]*\mathbb{E}_{t+1}[C_{t+3}^{ -\rho}]] =\mathbb{E}_{t+1}[C_{t+2}^{ -\rho}]\mathbb{E}_{t+1}[C_{t+3}^{ -\rho}] OR

4a-) \mathbb{E}_{t+1}[Z2_{t+1 }]=\mathbb{E}_{t+1}[C_{t+2}^{ -\rho}C_{t+3}^{ -\rho}] this i believe is wrong but again not %100 sure

  • But instead if i define the auxiliary variable as


5-) \mathbb{E}_{t}[Z1_{t}]=\mathbb{E}_{t} [C_{t}^{ -\rho}] expectations operators drop so Z1_{t}=C_{t}^{ -\rho} and

6-) Z1(+1) means Z1_{t+1}=C_{t+1}^{ -\rho} without expections

7-) Z2 means Z2_{t}=\mathbb{E}_{t} [C_{t}^{ -\rho}C_{t+1}^{ -\rho} ]

8 -) Z2(+1) means \mathbb{E}_{t+1}[Z2_{t+1 }]=\mathbb{E}_{t+1}[C_{t+1}^{ -\rho}C_{t+2}^{ -\rho}]

and the auxiliary variable Z1 enters without expectations operator because in the initial setup Z1=C^-(rho); is defined in a way that it holds in every period exactly

thanks a lot again and hopefully the way i described it is not confusing


You should have
Z2=E_t(Z1_tZ1_{t+1})=Z1_t E_t(E_{t+1}(C_{t+2}^{-\rho})))=E_t(C_{t+1}^{-\rho})E_{t}(C_{t+2}^{-\rho})
by the law of iterated expectations and the fact that Z1_t is contained in the time t information set.
I am not sure what you mean in point 8. Z2(+1) in Dynare is always E_t(Z2_{t+1}) as indicated by the +1 timing.

thanks a lot for the clarification. So as a general rule dealing with expectations
Z1=C^-(rho); ==> E_t[ Z1_{t}]= E_t [ (C_{t}^{-\rho})] ==>Z1_{t}={C_{t}^{-\rho}}
expectations cancel and this holds in every period so when i put it inside the auxiliary variable
Z2=Z1*Z1(+1); ==> E_t[ Z2]= E_t [ C_{t}^{-\rho}C_{t+1}^{-\rho}] and reasoning is

E_t[ Z2]= E_t [ C_{t}^{-\rho}\underbrace{C_{t+1}^{-\rho}}_{\small{because Z1(+1) holds \ exactly \ in \ every \ period \ without \ expectation }}]

i can substitute Z1_{t+n}={C_{t+n}^{-\rho}}, without worrying about expectations operator

but if i define Z1=C(+1)^-(rho); then this logic does not apply because Z1 is the expectation of a variable next period, not the exact value of that variable

Yes, exactly, although I did not get your intermediate step. I guess there is a typo on the LHS. It must be Z2 in the expectation.

sorry just corrected it. appreciate this very much. thanks a lot